Fluid Mechanics: Buoyancy, Pressure, and Fluid Flow

Posted on Aug 26, 2024 in Physics

5. Let the volume of the expanded air sacs be V_a and that of the fish with its air sacs collapsed be V. Then

where ρ_w is the density of the water. This implies

ρ_fishV = ρ_w(V + V_a) or (V + V_a)/V = 1.08/1.00,

which gives V_a/(V + V_a) = 0.074 = 7.4%.

7. (a) The pressure difference results in forces applied as shown in the figure. We consider a team of horses pulling to the right. To pull the sphere apart, the team must exert a force at least as great as the horizontal component of the total force determined by “summing” (actually, integrating) these force vectors.

We consider a force vector at angle θ. Its leftward component is Δp cos θdA, where dA is the area element for where the force is applied. We make use of the symmetry of the problem and let dA be that of a ring of constant θ on the surface. The radius of the ring is r = R sin θ, where R is the radius of the sphere. If the angular width of the ring is dθ, in radians, then its width is R dθ and its area is dA = 2πR² sin θ dθ. Thus the net horizontal component of the force of the air is given by

(b) We use 1 atm = 1.01 × 10⁵ Pa to show that Δp = 0.90 atm = 9.09 × 10⁴ Pa. The sphere radius is R = 0.30 m, so

F_h = π(0.30 m)²(9.09 × 10⁴ Pa) = 2.6 × 10⁴ N.

(c) One team of horses could be used if one half of the sphere is attached to a sturdy wall. The force of the wall on the sphere would balance the force of the horses.

13. In this case, Bernoulli’s equation reduces to Eq. 14-10. Thus,

17. The hydrostatic blood pressure is the gauge pressure in the column of blood between feet and brain. We calculate the gauge pressure using Eq. 14-7.

(a) The gauge pressure at the heart of the Argentinosaurus is

(b) The gauge pressure at the feet of the Argentinosaurus is

20. To find the pressure at the brain of the pilot, we note that the inward acceleration can be treated from the pilot’s reference frame as though it is an outward gravitational acceleration against which the heart must push the blood. Thus, with , we have

21. Letting p_a = p_b, we find

ρ_cg(6.0 km + 32 km + D) + ρ_m(y – D) = ρ_cg(32 km) + ρ_my

and obtain

27. (a) We use the expression for the variation of pressure with height in an incompressible fluid: p₂ = p₁ – ρg(y₂ – y₁). We take y₁ to be at the surface of Earth, where the pressure is p₁ = 1.01 × 10⁵ Pa, and y₂ to be at the top of the atmosphere, where the pressure is p₂ = 0. For this calculation, we take the density to be uniformly 1.3 kg/m³. Then,

(b) Let h be the height of the atmosphere. Now, since the density varies with altitude, we integrate

Assuming ρ = ρ₀ (1 – y/h), where ρ₀ is the density at Earth’s surface and g = 9.8 m/s² for 0 ≤ y ≤ h, the integral becomes

Since p₂ = 0, this implies

29. Eq. 14-13 combined with Eq. 5-8 and Eq. 7-21 (in absolute value) gives

mg = – kx (Ae/18Ae) .

With A₂ = 18A₁ (and the other values given in the problem) we find m = 8.50 kg.

38. If the alligator floats, by Archimedes’ principle the buoyancy force is equal to the alligator’s weight (see Eq. 14-17). Therefore,

                                                        .

If the mass is to increase by a small amount , then

.

With , the alligator sinks by

.

39. Let be the total volume of the iceberg. The non-visible portion is below water, and thus the volume of this portion is equal to the volume of the fluid displaced by the iceberg. The fraction of the iceberg that is visible is

                                                              .

Since the iceberg is floating, Eq. 14-18 applies:

Since , the above equation implies

                                                          .

Thus, the visible fraction is

(a) If the iceberg ( ) floats in saltwater with , then the fraction would be

.

(b) On the other hand, if the iceberg floats in fresh water ( ), then the fraction would be

.

40. (a) An object of the same density as the surrounding liquid (in which case the “object” could just be a packet of the liquid itself) is not going to accelerate up or down (and thus won’t gain any kinetic energy). Thus, the point corresponding to zero K in the graph must correspond to the case where the density of the object equals ρ_liquid. Therefore, ρ_ball = 1.5 g/cm³ (or 1500 kg/m³).

(b) Consider the ρ_liquid = 0 point (where K_gained = 1.6 J). In this case, the ball is falling through a perfect vacuum, so that v² = 2gh (see Eq. 2-16) which means that K = mv² = 1.6 J can be used to solve for the mass. We obtain m_ball = 4.082 kg. The volume of the ball is then given by m_ball/ρ_ball = 2.72 × 10^-3 m³.

41. For our estimate of V_submerged we interpret “almost completely submerged” to mean

Thus, equilibrium of forces (on the iron sphere) leads to

where r_i is the inner radius (half the inner diameter). Plugging in our estimate for V_submerged as well as the densities of water (1.0 g/cm³) and iron (7.87 g/cm³), we obtain the inner diameter:

47. (a) When the model is suspended (in air) the reading is F_g (its true weight, neglecting any buoyant effects caused by the air). When the model is submerged in water, the reading is lessened because of the buoyant force: F_g – F_b. We denote the difference in readings as Δm. Thus,

which leads to F_b = Δmg. Since F_b = ρ_wgV_m (the weight of water displaced by the model) we obtain

(b) The scaling factor is discussed in the problem (and for purposes of significant figures is treated as exact). The actual volume of the dinosaur is

(c) Using ρ ≈ ρ_w = 1000 kg/m³, we find

which yields 5.102 × 10³ kg for the T. rex mass.

51. This problem involves the use of the continuity equation (Eq. 14-23): .

(a) Initially, the flow speed is and the cross-sectional area is . At point a, as can be seen from Fig. 14-47, the cross-sectional area is

                                                           .

Thus, by the continuity equation, the speed at point a is

(b) Similarly, at point b, the cross-sectional area is , and therefore, by the continuity equation, the speed at point b is

53. Suppose that a mass Δm of water is pumped in time Δt. The pump increases the potential energy of the water by Δmgh, where h is the vertical distance through which it is lifted, and increases its kinetic energy by , where v is its final speed. The work it does is and its power is

Now the rate of mass flow is Δm/Δt = ρ_wAv, where ρ_w is the density of water and A is the area of the hose. The area of the hose is A = πr² = π(0.010 m)² = 3.14 × 10^–4 m² and

ρ_wAv = (1000 kg/m³) (3.14 × 10^–4 m²) (5.00 m/s) = 1.57 kg/s.

Thus,

63. (a) The friction force is

(b) The speed of water flowing out of the hole is v = . Thus, the volume of water flowing out of the pipe in t = 3.0 h is

65. (a) Since Sample Problem 14-8 deals with a similar situation, we use the final equation (labeled “Answer”) from it:

The stream of water emerges horizontally (θ₀ = 0° in the notation of Chapter 4), and setting y – y₀ = –(H – h) in Eq. 4-22, we obtain the “time-of-flight”

Using this in Eq. 4-21, where x₀ = 0 by choice of coordinate origin, we find

(b) The result of part (a) (which, when squared, reads x² = 4h(H – h)) is a quadratic equation for h once x and H are specified. Two solutions for h are therefore mathematically possible, but are they both physically possible? For instance, are both solutions positive and less than H? We employ the quadratic formula:

which permits us to see that both roots are physically possible, so long as x ≤ H. Labeling the larger root h₁ (where the plus sign is chosen) and the smaller root as h₂ (where the minus sign is chosen), then we note that their sum is simply

Thus, one root is related to the other (generically labeled h’ and h) by h’ = H – h. Its numerical value is

(c) We wish to maximize the function f = x² = 4h(H – h). We differentiate with respect to h and set equal to zero to obtain

or h = (40 cm)/2 = 20 cm, as the depth from which an emerging stream of water will travel the maximum horizontal distance.

67. (a) The continuity equation yields Av = aV, and Bernoulli’s equation yields , where Δp = p₁ – p₂. The first equation gives V = (A/a)v. We use this to substitute for V in the second equation, and obtain . We solve for v. The result is

(b) We substitute values to obtain

Consequently, the flow rate is

72. We use Bernoulli’s equation .

When the water level rises to height h₂, just on the verge of flooding, , the speed of water in pipe M, is given by

By the continuity equation, the corresponding rainfall rate is